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My course defines the level sets of a function $f:\mathbb{R}^n \to \mathbb{R}$ as

$$L_\alpha(f) = \{ x\in\mathbb{R}^n : f(x)=\alpha \}$$

For a problem, I need to find an $L_\alpha(f)$ that is a convex set. My issue is, how do I ever find a convex level set for a function that is not a straight line? For example the function $f(x,y)=x^2+y^2$ has a circular level set for all values of alpha.

tealing123
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  • What's the problem with a straight line? That's convex, isn't it? – Arthur Oct 07 '21 at 22:53
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    For $f(x,y)=x^2+y^2$ the level set for $\alpha=0$ is the single point $(0,0)$. And a single point in $\mathbb R^n$ is convex. – coffeemath Oct 07 '21 at 22:54
  • Yes a straight line is convex, but I am looking for other convex level sets – tealing123 Oct 07 '21 at 22:58
  • @coffeemath for any other value of $\alpha$ however, the level set is non-convex right? – tealing123 Oct 07 '21 at 23:00
  • $f(x) = \max(1, x^2 + y^2)$ has a level set $L_1(f) = {(x, y) \mid x^2 + y^2 \leq 1}$, which is a disc. – Mark Saving Oct 07 '21 at 23:06
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    @tealing123 Well when $\alpha<0$ the level set is empty, and so is vacuously convex. But for positive $\alpha$ the level set is not convex, as you say. – coffeemath Oct 07 '21 at 23:06
  • For an example in $\mathbb R^2$ where there are no level sets which are convex, consider $f(x,y)=y-x^2.$ Here the level set $f(x,y)=\alpha$ has as its graph $y=x^2+\alpha,$ which for any real $\alpha$ is a nondegenerate parabola. – coffeemath Oct 08 '21 at 04:44

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For a sufficiently smooth function $f$ "most" level sets can only be convex if they are hyperplanes or empty. Indeed, by Sard's theorem and implicit function theorem, for most values the level set is either empty or a closed submanifold of $\mathbb{R}^n$ of dimension $n-1$, and by implicit function theorem near any point such a level set is a graph over its tangent space. Such a graph can be convex only if it's a graph of a linear function, so the whole thing has to be linear. So you can get convex non-linear level sets only at "non-generic values" (but of course you can get them; any closed set in $\mathbb{R}^n$ is a level set of a smooth function).

Now, in convex analysis, what one usually looks at, is sublevel sets, $S_a(f)=\{x| f(x)\leq a\}$; those are going to be convex as soon as $f$ is.

Max
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  • Note that the sublevel sets being convex is its own convexity property of $f$, called "subconvex". There are non-convex functions with convex sublevel sets, like $-e^{-x^2}$. – Arthur Oct 08 '21 at 05:45
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    (Quasiconvex. The property is called quasiconvex, not subconvex.) – Arthur Oct 09 '21 at 21:52